","comment":{"@type":"Comment","text":" First find the value of angle $u$ and then find the value of hypotenuse using formula $\\operatorname{Cos}\\theta =\\dfrac{\\text{Adjacent side}}{\\text{Hypotenuse}}$ "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$10\\sqrt{2}cm$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$12\\sqrt{3}cm$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$15cm$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$14cm$$","position":2,"answerExplanation":{"@type":"Comment","text":"$$u{}^\\circ +2u{}^\\circ +90{}^\\circ =180{}^\\circ \\text{ }\\left[ \\text{Using Angle sum property} \\right] \\\\ \\Rightarrow u{}^\\circ =30{}^\\circ \\\\ \\text{Now, }\\operatorname{Cos}30{}^\\circ =\\dfrac{\\text{Adjacent side}}{\\text{Hypotenuse}} \\\\ \\Rightarrow \\dfrac{\\sqrt{3}}{2}=\\dfrac{7\\sqrt{3}}{\\text{Hypotenuse}} \\\\ \\Rightarrow \\text{Hypotenuse}=14cm $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Solving for a side in a right triangle Quiz 1","text":" If the hypotenuse of a right-angled triangle is $$4\\sqrt{29}cm$$ and the perpendicular is $$12cm$$ more than its base, then find the length of the base.","comment":{"@type":"Comment","text":" Assume base as $b$, then perpendicular as $b+12$. Now, solve for $b$ using Pythagoras theorem."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$6cm$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$10cm$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$11cm$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$8cm$$","position":1,"answerExplanation":{"@type":"Comment","text":"$$\\text{Let the base of a right-angled triangle be }b. \\\\ \\Rightarrow \\text{Perpendicular}=b+12 \\\\ \\text{Using Pythagoras theorem, we have:} \\\\ {{H}^{2}}={{P}^{2}}+{{B}^{2}} \\\\ \\Rightarrow {{\\left( 4\\sqrt{29} \\right)}^{2}}={{\\left( b+12 \\right)}^{2}}+{{\\left( b \\right)}^{2}} \\\\ \\Rightarrow 320=2{{b}^{2}}+24b \\\\ \\Rightarrow {{b}^{2}}+20b-8b-160=0 \\\\ \\Rightarrow b=8\\text{ }\\left( \\text{Neglect negative value} \\right) $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Solving for a side in a right triangle Quiz 1","text":" In the given figure if the sum of the two sides RS and ST is given by $$6a+150$$, then at what height is the parachute flying from the ground level? ","comment":{"@type":"Comment","text":" First find the value of $a$ using the given condition and then find required height using the formula $\\cos \\theta =\\dfrac{\\text{Adjacent side}}{\\text{Hypotenuse}}$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$150\\sqrt{3}\\text{ units}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$200\\sqrt{3}\\text{ units}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$500\\sqrt{3}\\text{ units}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$250\\sqrt{3}\\text{ units}$$","position":2,"answerExplanation":{"@type":"Comment","text":"$$5a+2a+50=6a+150 \\\\ \\Rightarrow 7a-6a=150-50 \\\\ \\Rightarrow a=100 \\\\ \\text{Now, }\\cos 30{}^\\circ =\\dfrac{\\text{Adjacent side}}{\\text{Hypotenuse}} \\\\ \\Rightarrow \\dfrac{\\sqrt{3}}{2}=\\dfrac{h}{5\\times 100} \\\\ \\Rightarrow h=250\\sqrt{3} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Solving for a side in a right triangle Quiz 1","text":" If $$x,y\\text{ and }z$$ are the lengths of the three sides $$RS,ST\\text{ and }RT$$ respectively of a right $$\\Delta RST$$ at $$\\angle T=90{}^\\circ $$ such that $$x:y:z=2:5:6$$ and $$\\cos R=\\dfrac{x}{32}$$, then find the value of $$z$$.","comment":{"@type":"Comment","text":" Assume $x,y\\text{ and }z\\text{ as }2t,5t\\text{ and }6t$ and then use the law of cosines formula to solve for $t$. Now, put the value of $t$ to get the required answer."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 30 units","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 42 units","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 54 units","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 60 units","position":3,"answerExplanation":{"@type":"Comment","text":"$$\\text{Let }x,y\\text{ and }z\\text{ be }2t,5t\\text{ and }6t\\text{ respectively}\\text{.} \\\\ \\text{Using law of cosines rule, we have:} \\\\ \\cos R=\\dfrac{R{{S}^{2}}+R{{T}^{2}}-S{{T}^{2}}}{2RS\\times RT} \\\\ \\Rightarrow \\dfrac{x}{32}=\\dfrac{{{x}^{2}}+{{z}^{2}}-{{y}^{2}}}{2x\\times z} \\\\ \\Rightarrow \\dfrac{2t}{32}=\\dfrac{4{{t}^{2}}+36{{t}^{2}}-25{{t}^{2}}}{2\\times 2t\\times 6t}\\Rightarrow \\dfrac{t}{16}=\\dfrac{15}{24} \\\\ \\Rightarrow t=10 \\\\ \\therefore z=6t=60\\text{ units} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Solving for a side in a right triangle Quiz 1","text":" $$XD$$ is the altitude of an isosceles $$\\Delta XYZ$$ such that $$XY=XZ$$. If $$a+b=7,XY=3a+2,YD=b+1\\text{ and }XZ=b+3$$, then find the length of $$XD$$ in $$cm$$.","comment":{"@type":"Comment","text":" Find the value of $a\\text{ and }b$ using $XY=XZ$ and the given condition, then find the value of $XY\\text{ and }YZ$. Now, solve for $XD$ by using Pythagoras theorem."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4\\sqrt{3}cm$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$7\\sqrt{5}cm$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$7\\sqrt{2}cm$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$6\\sqrt{3}cm$$","position":1,"answerExplanation":{"@type":"Comment","text":"
$$XY=XZ\\text{ }\\left[ \\because XYZ\\text{ is an isosceles triangle} \\right] \\\\ \\Rightarrow 3a+2=b+3 \\\\ \\Rightarrow 3a-b=1.........(i) \\\\ \\text{Given: }a+b=7......(ii) \\\\ \\Rightarrow a=2,b=5 \\\\ \\Rightarrow YD=b+1=6\\text{ and }XY=3(2)+2=12 \\\\ \\text{Now, }X{{Y}^{2}}=Y{{D}^{2}}+X{{D}^{2}}\\text{ }\\left[ \\text{Using Pythagoras theorem} \\right] \\\\ \\Rightarrow X{{D}^{2}}=144-36=108 \\\\ \\Rightarrow XD=6\\sqrt{3} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}